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Would a spin-2 particle necessarily have to be a graviton?

+ 21 like - 0 dislike
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I'm reading often that a possible reason to explain why the Nobel committee is coping out from making the physics Nobel related to the higgs could be among other things the fact that the spin of the new particle has not yet been definitively determined, it could still be 0 or 2.

This makes me wonder if the spin would (very very surprisingly!) finally be discovered to be 2, this then necessarily would mean that the particle has to be a graviton? Or could there hypothetically be other spin-2 particles? If not, why not and if there indeed exist other possibilities what would they be?

asked Oct 10, 2012 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
I don't think so, since the observed particle is massive and the graviton is massless (unless there is some mechanism ala Higgs which gives it mass). If it was spin 2, it would be much more likely that its a bound states of some particles rather than the graviton. But I don't think you can construct any spin 2 bound states of standard model particles, such as quarks, with so low mass.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Heidar
I think there is a pretty good argument in Weinberg that no other fundamental particle can have spin-2 besides the graviton (I can't recall it atm), but composite particles can certainly have spin-2, which I think is what is one possible exciting result if Higgs is spin-2, but that is so far away...

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user kηives

3 Answers

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There are theoretical arguments that a massless spin-2 particle has to be a graviton. The basic idea is that massless particles have to couple to conserved currents, and the only available one is the stress-energy tensor, which is the source for gravity. See this answer for more detail.

However, the particle discovered at LHC this year has a mass of 125 GeV, so none of these arguments apply. It would be a great surprise if this particle did not have spin 0. But it is theoretically possible. One can get massive spin 2 particles as bound states, or in theories with infinite towers of higher spin particles.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user user1504
answered Oct 10, 2012 by user1504 (1,100 points) [ no revision ]
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Only if the dark matter sector violates the assumptions of the Coleman-Mandula theorem, which would be very weird indeed.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user user1504
does spin-2 inherently require a mixing of spacetime degrees of freedom with particle degrees of freedom? I'm just imagining some sort of dark matter model that has a conserved tensor current of dark matter that is not necessarily its stress-energy tensor. It would be exotic, sure, but I don't think necessarily impossible.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Jerry Schirmer
@JerrySchirmer: It's not that easy to construct conserved tensor currents in theories with only massive particles. See the linked answer.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user user1504
Comment by anonymous user: the stress-energy tensor is not the only tensor which could couple to a spin-2 particle, the electromegnetic field tensor also could couple and photons are yet massless

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Neuneck
To the anonymous editor reverted by Neuneck: The field strength F decomposes as a sum of $(1,0)$ and $(0,1)$ representations. This is not what most people mean by spin 2.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user user1504
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Darn, Lenny Susskind has explained the Regge trajectories and I have just forgotten about it (stupid me!), so thanks for the reminder ... :-)

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Dilaton
Other examples are the Vasiliev "higher spin gauge theories".

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Mitchell Porter
+ 6 like - 0 dislike

It depends on your definition of "particle" . In the particle data group listings there exist a number of spin 2 resonances. These ultimately will be built up by quarks.

f_2(1270 MeV) page 9

a_2(1320) page 11

etc

there is a pi_2 (1670) page 16

So the bump called now "the Higgs" could turn out to be one more resonance. Not the graviton as it is envisaged in possible theories of everything, since gravity is long range and it should be massless, I believe.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user anna v
answered Oct 10, 2012 by anna v (1,710 points) [ no revision ]
Dear Anna, would it be possible for such a compound spin-2 "particle" with a mass of 125 GeV to be produced by the same processes, decay into the same channels, and generelly behave in the same manner (similar to a SM higgs) is it is measured so far at the LHC ? Would spin 2 of the measured particle mean that we have to look for something else that breaks the EW symmetry or could ist still do the job? Sorry if I'm confused maybe I should just shut up and read the link you gave to me :-)

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Dilaton
@Dilaton To identify it as a Higgs, they would have to nail the spin as zero, measure the branching ratios and find them consistent with the values predicted by the SM. If the spin turns out to be 2 we will have to look elsewhere for the Higgs. A spin 2 cannot do the job of the Higgs. Then the game is open. It will certainly still be very interesting to see how the phenomenologists deal with a spin 2 bump at 125GeV. Personally I think it will turn out to be the Higgs, but we have to wait for the statistics to say so unequivocally.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user anna v
Thanks Anna, I rather think it is the higgs too ...

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Dilaton
+ 6 like - 0 dislike

A massive spin 2 particle must have five modes: helicity $\pm 2$, $\pm 1$, 0. If a massless spin 2 particle has only helicity $\pm 2$ modes without other modes and has a dispersion $\omega = c k$, then such a massless spin 2 particle must be graviton (at least at linear order).

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Xiao-Gang Wen
answered Oct 11, 2012 by Xiao-Gang Wen (3,309 points) [ no revision ]
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... this reminds me slithly of the argument about why the photon has only two polarizations, is it an extension of this ...?

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Dilaton
No. It is the reverse. If a particle has only helicity $\pm 1$ modes without other modes and has a dispersion $\omega=ck$, then it has to be massless photon. But a massless particle with $\omega=ck$ can have helicity $\pm 1$, 0 modes. The phonon in 3D crystal is such massless particle.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Xiao-Gang Wen
@Xiao-GangWen That is because you do not have Poincaré invariance in condensed matter. In a relativistic theory, massless particles can only have helicity $0$, $\pm 1$ or $\pm 2$. If you allow Lorentz breaking at high energies then it is ok.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user drake
Even in a relativistic theory, the low energy excitations can be helicity $0$, $\pm 1$, AND $\pm 2$. You may call it to have three types of particles. A crystal do have helicity $0$ AND $\pm 1$ low energy excitations.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user Xiao-Gang Wen
@Xiao-GangWen Is there any symmetry that connects those excitations? The reason why photons have two polarizations —rather than being two independent particles— is that photons only participate in electromagnetic interactions —which are invariant under parity transformations— and parity connects the +1 helicity with the -1 helicity.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user drake
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(cont.) much like massless neutrinos: neutrinos have helicity $-1/2$ and antineutrinos helicity $+1/2$. This is due to neutrinos interact weakly and weak interactions do not preserve parity.

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user drake
Sorry: where it says "weak interactions —which are parity invariant—", it should obviously say "weak interactions —which are not parity invariant—"

This post imported from StackExchange Physics at 2014-03-09 15:49 (UCT), posted by SE-user drake

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